Inplicit Differentiation Problem

**amey** · November 19th 2011, 02:53 AM

The problem is to prove that $\frac{dy}{dx}=\frac{-1}{(x+1)^2}$ when $x\sqrt{y+1}+y\sqrt{x+1}=0$ . I tried substituting $x=\tan^2\theta$ and $y=\tan^2\phi$ but then I got stuck. I tried many other things but this was the only notable one.

thanks.

**sbhatnagar** · November 19th 2011, 04:00 AM

Do you know about the chain rule?

For example: Solve $\frac{d}{dx}(\sqrt{x+1})$

Let $y=\sqrt{x+1}$ and $u=x+1$

Then: $\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}=(1)(\frac {1}{2\sqrt{u}})=\frac{1}{2\sqrt{x+1}}$

**amey** · November 19th 2011, 04:04 AM

Yes, I know the basic formulae and algebra of derivatives.

**tom@ballooncalculus** · November 19th 2011, 04:53 AM

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

Larger

Solve the bottom row (for dy/dx).

Edit: ... which is easier said than done. Wolfram doesn't have any better ideas: differentiate x sqrt(y+1) + y sqrt(x+1) = 0 with respect to x - Wolfram|Alpha.

Ok, I'm beginning to see the tan^2 idea...

... nah! Was it suggested? What is the context?

**skeeter** · November 19th 2011, 07:12 AM

I took the original equation, moved one term to the RHS, squared both sides of the resulting equation and arrived at this quadratic equation ...

$(x+1)y^2 - x^2y - x^2 = 0$

using the quadratic formula ...

$y = \frac{x^2 \pm \sqrt{x^4 - 4(x+1)(-x^2)}}{2(x+1)}$

which simplifies to either $y = x$ (which, by inspection, does not work in the original equation) or $y = \frac{-x}{x+1}$

the derivative of the second equation is $y' = \frac{-1}{(x+1)^2}$

might be some other way, but I don't see it ...

**FernandoRevilla** · November 19th 2011, 07:22 AM

Originally Posted by skeeter

might be some other way, but I don't see it ...

You have chosen the best way in this case.

**amey** · November 19th 2011, 04:26 PM

I agree with FernandoRevilla. this is the best way. thanks again.

Thread: Inplicit Differentiation Problem

LinkBack

Thread Tools

Display

Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Re: Inplicit Differentiation Problem

Similar Math Help Forum Discussions

Differentiation problem

Differentiation Problem

Differentiation Problem

Differentiation problem

differentiation problem #3

Search Tags